Question

A manufacturer of power tools believes that the mean amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The management is planning to use the assembly time in advertising and promotion of the product and the competitors have an average of 90 minutes assembly time. That’s why they are interested in determining the probability that the sample mean will be between 77 and 89 minutes, less than 82 minutes, and greater than 88 minutes. Based on the analysis what would be your suggestion to the management.

Answer 1

Solution :

Given that ,

mean = = 80

standard deviation = = 40

n = 64

=    = 80

= / n = 40 / 64 = 5

a) P(77 < < 89)  

= P[(77 - 80) / 5 < ( - ) / < (89 - 80 ) / 5 )]

= P( -0.60 < Z < 1.80 )

= P(Z < 1.80 ) - P(Z < -0.60 )

Using z table,  

= 0.9641 - 0.2743  

= 0.6898

b) P( < 82) = P(( - ) / < (82 - 80) / 5)

= P(z < 0.40)

Using z table

= 0.6554

c) P( > 88) = 1 - P( < 88)

= 1 - P[( - ) / < (88 - 80) / 5]

= 1 - P(z < 1.60)

Using z table,    

= 1 - 0.9452

= 0.0548